Let $\Omega \subset \mathbb R^2$ be a bounded Lipschitz domain (lets say $\Omega = (0,1)^2$). Let $u \colon \Omega \longrightarrow \mathbb R$ be a function in the Sobolev space $H^1(\Omega)$, that is $u$ is weakly differentiable and $u, |\nabla u| \in L^2(\Omega)$. Additionally, assume $0 \leq u \leq 1$.
Can I conclude, that $u$ is continuous?
I know that merely $u \in H^1(\Omega)$ does not provide $u \in C^0(\Omega)$ e.g. $u(x) = \log \log |x|$. But the continuity only fails to hold because of the singularity in $x = 0$, which is removed when I assume $u$ to be bounded.
Because the dimension of $\Omega$ and the integrability and differentiability of $H^1$ are critical, I was not able to find a result which is connected to imbedding theorems.
I would be glad if someone could clarify my question with literature, or a counterexample.
Thanks in advance.
Counterexample: $f(x)=\sin^2 \log\log \frac1{|x|}\in H^1\cap L^\infty$, $0\le f \le 1$, and is not continuous at $0\in \Omega:= B_{1/2}(0)$.
This is because as $x\to0$, $\log\log\frac1{|x|}\to\infty$, so $\sin^2\log\log\frac1{|x|}$ takes all values in $[0,1]$ on any neighbourhood of $0$.
That $f\in H^1$ is a consequence of Exercise 17 of Evans Chapter 5 (2nd ed.), or by directly computing the weak derivative; set $\operatorname{LL}(x)=\log\log\frac1{|x|}$. Then the $\alpha$th-derivative everywhere except 0, $|\alpha|=1$, is $$ g_\alpha:= 2(\sin \operatorname{LL} \cos \operatorname{LL} ) D^\alpha \operatorname{LL} \in L^2$$ and for $\phi\in C_c^\infty(\Omega)$, $$ \int_{\epsilon<|x|<1/2} f D^\alpha \phi = \int_{|x|=\epsilon} f\phi n^\alpha d\sigma(x) -\int_{\epsilon<|x|<1/2} g_\alpha \phi$$ sending $\epsilon\to 0$ shows $D^\alpha f = g_\alpha$ in the sense of weak derivatives.